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Derive the relationship between relative lowering of vapour pressure and mole fraction ofthe volatile liquid. |
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Answer» Solution : Let A and B represent the solvent and the SOLUTE respectively. `x_A`and `x_B`stand for the molefraction of solvent and solute, respectively. `p_A^0` and `P_A` represent the vapour pressure of A in the pure liquid and solution, respectively. `p_A prop x_A` `p_A = p_A^0 xx x_A`....(i) `x_A + x_B = 1` form (i) and (ii) we have `p_A = p_A^0 ( 1- x_B)` `(P_A)/(p_A^0) = 1 - x_B " or" x_B= 1 - (p_A)/(p_A^0) = (p_A - p_A^0)/(p_A^0) " or " (p_A^0 -p_A)/(p_A^0) = x_B` RELATIVE lowering of vapour pressure of a solution is equal to mole FRACTION of the solute. |
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